Answer:
K2SO4, NH3, HOCI, HCI, CH3NH2, SiCl4, CO2, CH20
Explanation:
Substances are soluble in water when they are ionic or polar covalent substances.
If we look at the substances listed, K2SO4 is ionic while NH3, HOCI, HCI, CH3NH2, SiCl4, CO2, CH20 all contain polar covalent bonds which accounts for their water solubility.
Hence ionic and polar covalent substances are soluble in a polar solvent such as water.
Answer:- 10 L of ethane.
Solution:- The given balanced equation is:

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.
Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

= 
From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L
Hence, 10 L of ethane will be remaining.
Answer:
no but that is what we were taught
Answer:
10.9%.
Explanation:
The first thing to do in order to solve this question is to Determine the value for the volume of the the cube. This can be done by taking the cube root of the length of the cube;
The volume of the cube = (length of the cube)^3 = length × length × length = 1.72 × 1.72 × 1.72 =( 1.72)^3 = 5.09cm^3.
The next thing you do is to Determine the exponential density, the can be done by using the formula below;
The exponential density = mass/ volume = 55. 786/ 5.09 = 10.96 g/cm^3.
Therefore, the percent error = (true density of the cube - exponential density of the cube)÷ true density of the cube × 100.
Hence, the percent error = 12.30 - 10.96/12.30 × 100 = 10.9%.
Answer:
the correct answer is Blue